Question: Solve for $x$ and $y$ using elimination. $\begin{align*}8x-2y &= 8 \\ x+y &= 2\end{align*}$
Answer: We can eliminate $y$ when its corresponding coefficients are negative inverses. Recalling our knowledge of least common multiples, multiply the top equation by $1$ and the bottom equation by $2$ $\begin{align*}8x-2y &= 8\\ 2x+2y &= 4\end{align*}$ Add the top and bottom equations. $10x = 12$ Divide both sides by $10$ and reduce as necessary. $x = \dfrac{6}{5}$ Substitute $\dfrac{6}{5}$ for $x$ in the top equation. $8( \dfrac{6}{5})-2y = 8$ $\dfrac{48}{5}-2y = 8$ $-2y = -\dfrac{8}{5}$ $y = \dfrac{4}{5}$ The solution is $\enspace x = \dfrac{6}{5}, \enspace y = \dfrac{4}{5}$.